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# 第六部分 电缆的 EMC 设计 - PowerPoint PPT Presentation

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### 第六部分电缆的EMC设计

• 场在导线中感应的噪声
• 电缆之间的串扰

1V/m场强产生的电压

dBV

3

1

2

0

-10

h = 0.5m

S: A = 100m

B = 30m

C = 10m

D = 3m

E = 1m

A

-20

B

C

-30

D

-40

E

-50

10kHz 100kHz 1MHz 10MHz 100MHz 1GHz 10GHz

VD

V1

I1

V2

I2

VC

CMRR = 20lg ( VC / VD )

CMRR

CMRR

f

f

0V

VN

VN＝ ( d  / dt ) = A ( dB / dt )

VS

VS

VS

100

(A)

1M

0

13

100

(D)

1M

27

100

(B)

1M

28

100

1M

(E)

1M

13

100

(C)

100

80

(F)

1M

63

100

(I)

1M

55

100

1M

(G)

77

1M

70

100

100

(J)

1M

(H)

R0

RL

C

M

IL

R2G

R2L

IL

IC

IC

ZSZL < 3002： 磁场耦合为主

ZSZL > 10002： 电场耦合为主

3002 < ZSZL < 10002：取决于几何结构和频率

C12

C12

V1

VN

C1G

C2G

C2G

R

C1G

V1

R

j  [ C12 / ( C12 + C2G)]

VN ＝

V1

j  + 1 / R ( C12 + C2G)]

j  [ C12 / ( C12 + C2G)]

VN ＝

V1

j  + 1 / R ( C12 + C2G)]

R >> 1 / [ j  ( C12 + C2G )]

R << 1 / [ j  ( C12 + C2G )]

VN = V1 [ C12 / ( C12 + C2G ) ]

VN = j R C12 V1

VN = j RC12V1

C12V1

VN =

(C12 + C2G)

1 / R (C12 + C2G)

C2S

C1s

C1s

V1

C1G

CSG

Vs

C1G

V1

Vs

CsG

C12

C2S

C1s

C1s

C1G

C12

CSG

C2G

V1

VN

VN

V1

CsG

R 很大时：VN = V1 [ C12 / ( C12 + C2G + C2S ) ]

R 很小时：VN = jRC12

a

b

a

 1是电流I1在回路1中产生的磁通， 12 是电流I1在回路2中产生的磁通

M = （  / 2  ）ln[b2/（b2- a2）]

I1

R2

I1

R1

M

VN

V1

VN

V1

R2

R

R1

R

VN ＝ d12 / dt = d(MI1)/dt = M dI1 / dt

IN = j  C12V1

V

R1

R2

VN = j  M12 I1

V

R1

R2

I1

M1S

M12

I1

~

V12 = j  M12 I1

VS2 = j  MS2 IS

VN = V12 + VS2

M1S

M12

IS

MS2

+ -

- +

V12

VS2

VS2项求解

LS = / ISMS2 = / IS

+

+

+

+

VS2 = j MS2 I S

= jMS2 ( V S / ZS)

= jLS [ V S / ( jLS+RS )]

= VS [ j / ( j+RS/LS)]

+

+

+

+

+

VN = V12 + VS2

V12 = j M12I1 VS = j M1SI1

V12

VN = V12 - V12[ j / ( j+RS / LS)]

= V12 [ (RS / LS) / ( j+RS / LS)]

VN

VN＝ M12 I1（Rs / Ls )

VN＝ j M12 I1

lg

Rs / Ls

/10 /4 /2 3/4  lg f